how to get right angle, or add these two complex numbers?

[kraz]

Hi, all, i'm struggling to solve this complex number.

[math]((\frac{\sqrt 3}{2}+\frac{i}{2})^{2021}+(\frac{\sqrt 3}{2}-\frac{i}{2})^{2021})[/math]

I don't know how to go with this one in my math workbook (college)

I don't know how i can add them (like they are usually added, real with real, and imaginary with imaginary), if i add them usual way, i get, Imaginary part to be 0
But when i insert this problem in calculator, i get -1.73
Also, this big square (2021) is confusing me even more)

I tried to put it in polar form, and somehow to get rid of 2021 first, but, i can't finish it, because i get very weird numbers (or is it's because i put them wrong way in calculator?)

When trying to put it in polar form, only this i can get right is intensity, which is 1, but i dont know how to get angle.
I tried to get angle with putting [math]\frac{cos \frac{1}{2}}{sin \frac{\sqrt 3}{2}}[/math] because [math]tan^{-1}[/math] is [math]\frac{cos}{sin}[/math]And i get [math]\frac{1}{0.021}[/math] which i then calculate and get 46.78 degrees

But that look wrong. Because, on calculator i got -1.73
That doesn't look like good way to get that angle.

So, my biggest problem and question here, is, how to get angle right, and how to sum them (maybe, polar form is not needed?) ?

Because, this 2021, i think it's done, by simplifyning, because it goes into circles . it would be 5 i think.
But without getting angle right way, or, how to add them at all, i can't solve this at all.

 

[Steven G]

(cos t + i sin t)²⁰²¹ = (cos 2021t + i sin 2021t)

You need to write [math](\frac{\sqrt 3}{2}+\frac{i}{2})[/math] in the form (cos t + i sin t) and then use the formula above.

Also tan^-1 is not sin/cos

 

[kraz]

(cos t + i sin t)²⁰²¹ = (cos 2021t + i sin 2021t)

You need to write [math](\frac{\sqrt 3}{2}+\frac{i}{2})[/math] in the form (cos t + i sin t) and then use the formula above.

Also tan^-1 is not sin/cos

Okay, i tried: to do this: [math]tg^{-1}\frac{\frac{1}{2}}{\frac{\sqrt 3}{2}}[/math]then i get: [math]tg^{-1}\frac{2\sqrt 3}{2}[/math]which i then: [math]tg^{-1}\sqrt 3[/math]and when i get tat inverse tangent of [imath]\sqrt3[/imath] , i get 60 degrees.

i then put that into polar form
i get:

[math]1*(cos 60° + i sin60°)[/math]
and then to apply tha 2021 exponent:

[math]1*(cos 2021* 60° + i sin 2021* 60°)[/math]
it would be too big number.

And also, how i would sum it with another number if it's in polar form?

Because when i add them directly in rectangular form, i get real part to be [imath]\sqrt 3[/imath] but imaginary part becomes 0? because [math]\frac{i}{2}+(-\frac{i}{2}) = 0?[/math]
what i'm mising here?

 

[Dr.Peterson]

I tried to get angle with putting [imath]\frac{cos \frac{1}{2}}{sin \frac{\sqrt 3}{2}}[/imath] because [imath]tan^{-1}[/imath] is [imath]\frac{cos}{sin}[/imath]
And i get [imath]\frac{1}{0.021}[/imath] which i then calculate and get 46.78 degrees

You are being confused by the notation. [imath]tan^{-1}[/imath] does not mean the reciprocal of the tangent (which is the cotangent), but the inverse tangent function, also called "arctan".

Have you never been taught how to find the angle that has a given tangent, by using this button on your calculator? For information on that, see

Inverse Sine, Cosine, Tangent

www.mathsisfun.com

On the other hand, this is a special angle that you are probably expected to recognize immediately.

 

[pka]

solve this complex number.
[imath](\frac{\sqrt 3}{2}+\frac{i}{2})^{2021}+(\frac{\sqrt 3}{2}-\frac{i}{2})^{2021})[/imath]

This is somewhat of a trick question.
If [imath] z=\left(\dfrac{\sqrt3}{2}+\dfrac{\bf i}{2}\right)[/imath] then [imath]r=|z|=1[/imath] and [imath] \arg(z)=\dfrac{\pi}{6}[/imath]
So that [imath]z={\large {\bf 1}}\exp\left(\dfrac{i\pi}{6}\right)[/imath] Thus what is [imath]z^{2021}=~?[/imath]
Moreover, if [imath] z=\left(\dfrac{\sqrt3}{2}-\dfrac{\bf i}{2}\right)[/imath] then [imath]r=|z|=1[/imath] and [imath] \arg(z)=-\dfrac{\pi}{6}[/imath]

 

[Steven G]

Cos t + i sin t has t in radians, not degrees. Also, 60^o is not even the correct angle anyways.
Try again.

 

[Steven G]

Okay, i tried: to do this: [math]tg^{-1}\frac{\frac{1}{2}}{\frac{\sqrt 3}{2}}[/math]then i get: [math]tg^{-1}\frac{2\sqrt 3}{2}[/math]which i then: [math]tg^{-1}\sqrt 3[/math]and when i get tat inverse tangent of [imath]\sqrt3[/imath] , i get 60 degrees.

i then put that into polar form
i get:

[math]1*(cos 60° + i sin60°)[/math]
and then to apply tha 2021 exponent:

[math]1*(cos 2021* 60° + i sin 2021* 60°)[/math]
it would be too big number.

And also, how i would sum it with another number if it's in polar form?

Because when i add them directly in rectangular form, i get real part to be [imath]\sqrt 3[/imath] but imaginary part becomes 0? because [math]\frac{i}{2}+(-\frac{i}{2}) = 0?[/math]
what i'm mising here?

What would be too big a number? Cos (2021* 60°) will be large? You do know that Cos t varies between -1 and 1, correct?

 

[Dr.Peterson]

Okay, i tried: to do this: [math]tg^{-1}\frac{\frac{1}{2}}{\frac{\sqrt 3}{2}}[/math]then i get: [math]tg^{-1}\frac{2\sqrt 3}{2}[/math]which i then: [math]tg^{-1}\sqrt 3[/math]and when i get tat inverse tangent of [imath]\sqrt3[/imath] , i get 60 degrees.

One error is in simplifying [imath]\frac{\frac{1}{2}}{\frac{\sqrt 3}{2}}[/imath]. Check your work; it is not [imath]\sqrt 3[/imath]. Fix that, and you'll get the correct angle.

And also, how i would sum it with another number if it's in polar form?

Because when i add them directly in rectangular form, i get real part to be [imath]\sqrt 3[/imath] but imaginary part becomes 0? because [math]\frac{i}{2}+(-\frac{i}{2}) = 0?[/math]
what i'm mising here?

You'll be able to put it back in rectangular form very easily!

And if you think carefully about the problem, you'll see from the start that you are adding two conjugates (because powers of conjugates are still conjugates), which always results in a real number.

But how are you saying you added them? Please show work.

 

[kraz]

And if you think carefully about the problem, you'll see from the start that you are adding two conjugates (because powers of conjugates are still conjugates), which always results in a real number.

can you explain this further to me. maybe this can help me to understand it completely.

so, far, i have managed to make something out of this, but as i said, this + betwen them is confusing me what it means, i need to do with it.


I found out from unit circle that it is indeed [imath]\frac{π}{6}[/imath] .
and 2021, is in fact power to 5, when i remove those +2π as it goes around circle.

i applied power of 5 to this polar form i managed to get
[math]1^5 (cos 5* \frac{π}{6} + i sin 5* \frac{π}{6} )[/math]
and so i got
[math]1 (cos \frac{5π}{6} + i sin \frac{5π}{6} )[/math]
and all i have left to do that is confusing me, is what to do with that conjugate.
if i calculate that conjugate, i would get same this result, but just with opposite sign.
and if i add them together ( because of + sign between them) , it would be 0

so, only that sign + confuses me.
what i should do with this i got?

 

[Steven G]

5^1=5
5^2=25
5^3=125
5^4=625
5^5=3125
5^6=15625
Do you anything in common with all the numbers above? Yes, they all end in 5. This will be true for all positive integer powers of 5
So how can an integer power of 5 equal 2021?

 

[Dr.Peterson]

can you explain this further to me. maybe this can help me to understand it completely.

so, far, i have managed to make something out of this, but as i said, this + betwen them is confusing me what it means, i need to do with it.


I found out from unit circle that it is indeed [imath]\frac{π}{6}[/imath] .
and 2021, is in fact power to 5, when i remove those +2π as it goes around circle.

i applied power of 5 to this polar form i managed to get
[imath]1^5 (cos 5* \frac{π}{6} + i sin 5* \frac{π}{6} )[/imath]

and so i got
[imath]1 (cos \frac{5π}{6} + i sin \frac{5π}{6} )[/imath]

and all i have left to do that is confusing me, is what to do with that conjugate.
if i calculate that conjugate, i would get same this result, but just with opposite sign.
and if i add them together ( because of + sign between them) , it would be 0

so, only that sign + confuses me.
what i should do with this i got?

If you're saying what you should be saying (I'm not quite sure), you have found the first term, [imath]\cos \frac{5π}{6} + i \sin \frac{5π}{6}[/imath], and you see that the second term is its conjugate, [imath]\cos \frac{5π}{6} - i \sin \frac{5π}{6}[/imath].

When you add these together, the imaginary part becomes zero, but not the real part, right? More generally, [imath](a+ib)+(a-ib)=2a+i0 = 2a[/imath].

 

[Dr.Peterson]

I found out from unit circle that it is indeed π6\frac{π}{6}6π .
and 2021, is in fact power to 5, when i remove those +2π as it goes around circle.

I think Jomo misunderstood what you are saying here. What you mean (but didn't clearly say) is that raising this particular complex number to the 2021st power is equivalent to raising it to the 5th power, because its 12th power is 1, and 2021 = 12*168+5.

 

[kraz]

I think Jomo misunderstood what you are saying here. What you mean (but didn't clearly say) is that raising this particular complex number to the 2021st power is equivalent to raising it to the 5th power, because its 12th power is 1, and 2021 = 12*168+5.

yes, i meant that

 

[kraz]

If you're saying what you should be saying (I'm not quite sure), you have found the first term, [imath]\cos \frac{5π}{6} + i \sin \frac{5π}{6}[/imath], and you see that the second term is its conjugate, [imath]\cos \frac{5π}{6} - i \sin \frac{5π}{6}[/imath].

When you add these together, the imaginary part becomes zero, but not the real part, right? More generally, [imath](a+ib)+(a-ib)=2a+i0 = 2a[/imath].

oh it is supposed to happen ?
------------
So far, i got clearly understanding of this, and got the same value as in calculator.


Problem:

[math]((\frac{\sqrt 3}{2}+\frac{i}{2})^{2021}+((\frac{\sqrt 3}{2}-\frac{i}{2})^{2021}[/math]


i solved as:


putting it in polar coordinates which are (radian degrees)

[math]1(cos\frac{π}{6}+isin\frac{π}{6})[/math]

applying 2021 exponent, but in fact applying exponent of 5 because , because its 12th power is 1, and 2021 = 12*168+5.

[math]1^5(cos5*\frac{π}{6}+isin5*\frac{π}{6})[/math]

getting

[math]1(cos\frac{5π}{6}+isin\frac{5π}{6})[/math]

Bringing it back in rectangular form, to add it with another complex number.

And that another complex number is just this complex number conjugate.


So


[math](-\frac{\sqrt 3}{2} + i\frac{1}{2}) + (-\frac{\sqrt 3}{2} - i\frac{1}{2})[/math]

i would get


complex part

[math]( i\frac{1}{2}) + ( - i\frac{1}{2}) = i0[/math]

real part

[math](-\frac{\sqrt 3}{2} ) + (-\frac{\sqrt 3}{2} ) = (-\frac{\sqrt 3}{2} ) -(\frac{\sqrt 3}{2} ) = -\frac{2\sqrt 3}{2} = - \sqrt 3[/math]

and there we have it, the result is [imath]- \sqrt 3[/imath]


and [math]-\sqrt 3 = - 1.73205...[/math]
which is exactly value i get in calculator


So, this is supposed to happen, and this is how it's done?

 

[Dr.Peterson]

So, this is supposed to happen, and this is how it's done?

Yes. The fact that the sum of a number and its conjugate is always real, which is obvious when you graph that sum on the complex plane, is a fact worth knowing!